Among `[Ni(CO)_4], [Ni(CN)_4]^(2-), [NiCl_4]^(2-)` species, the hybridization states at the `Ni` atom are, respectively (At. no.of `Ni=28`)

Options

A. (a) `sp^3 , sp^3, dsp^2`
B. (b) `dsp^2, sp^3, sp^3`
C. (c) `sp^3, dsp^2, dsp^2`
D. (d) `sp^3, dsp^2, sp^3`

Explanation

To determine the hybridization states of the nickel atom in the complexes \([Ni(CO)_4]\), \([Ni(CN)_4]^{2-}\), and \([NiCl_4]^{2-}\), we will analyze each complex step by step.

### Step 1: Analyze \([Ni(CO)_4]\)

1. **Oxidation State**: In \([Ni(CO)_4]\), CO is a neutral ligand, so the oxidation state of Ni is 0.
2. **Electron Configuration**: The electron configuration of Ni (atomic number 28) is \([Ar] 3d^8 4s^2\).
3. **Hybridization**: Since CO is a strong field ligand, it will cause pairing of electrons in the 3d subshell. The configuration after pairing will be \(3d^{10}\) and \(4s^0\).
4. **Hybridization Type**: The four ligands (CO) will arrange themselves in a tetrahedral geometry, leading to \(sp^3\) hybridization.

### Step 2: Analyze \([Ni(CN)_4]^{2-}\)

1. **Oxidation State**: In \([Ni(CN)_4]^{2-}\), since CN is a -1 charge ligand, the oxidation state of Ni can be calculated as follows:
- Let the oxidation state of Ni be \(x\).
- \(x + 4(-1) = -2\) (charge of the complex)
- Thus, \(x - 4 = -2 \Rightarrow x = +2\).
2. **Electron Configuration**: The electron configuration of Ni in the +2 state is \(3d^8 4s^0\).
3. **Hybridization**: CN is also a strong field ligand, which will cause pairing of electrons. The configuration remains \(3d^8\) and \(4s^0\).
4. **Hybridization Type**: The geometry is square planar, leading to \(dsp^2\) hybridization.

### Step 3: Analyze \([NiCl_4]^{2-}\)

1. **Oxidation State**: In \([NiCl_4]^{2-}\), Cl is a -1 charge ligand, so the oxidation state of Ni can be calculated as follows:
- Let the oxidation state of Ni be \(x\).
- \(x + 4(-1) = -2\) (charge of the complex)
- Thus, \(x - 4 = -2 \Rightarrow x = +2\).
2. **Electron Configuration**: The electron configuration of Ni in the +2 state is \(3d^8 4s^0\).
3. **Hybridization**: Cl is a weak field ligand, which means it does not cause pairing of electrons. Therefore, the configuration remains \(3d^8\) with \(4s^0\) and \(4p^0\).
4. **Hybridization Type**: The geometry is tetrahedral, leading to \(dsp^3\) hybridization.

### Summary of Hybridizations
- For \([Ni(CO)_4]\): \(sp^3\)
- For \([Ni(CN)_4]^{2-}\): \(dsp^2\)
- For \([NiCl_4]^{2-}\): \(dsp^3\)

### Final Answer
The hybridization states at the Ni atom are, respectively:
- \([Ni(CO)_4]\): \(sp^3\)
- \([Ni(CN)_4]^{2-}\): \(dsp^2\)
- \([NiCl_4]^{2-}\): \(dsp^3\)